Need help understanding the logical structure behind solving an equation?

Question

Suppose I have an equation $f(x) = 0$, and I know the solution exists. The solution is $x = a$ (I don't know the actual solution to begin with, I just know it exists). I perform a series of $irreversible$ operations and arrive at the answer:

$f(x) = 0 \implies \cdots = \cdots \implies \cdots = \cdots \implies x = a$

which is just a big $f(x) = 0 \implies x = a$.

Since the operations are irreversible, we cannot go in the other direction and claim $ x = a \implies f(x) = 0$

Nevertheless, we know that the solution is $a$, so if we plug in $a$ into $f$ we get zero. But hold on a minute, the statement "if we plug in $a$ into $f$ we get zero" really means $x = a \implies f(x) = 0$; we just went in the other direction, which is something we said we couldn't do!

What is wrong here? Does the fact that $f(x) = a$ has a solution imply that there must be a reversible series of steps between $f(x) = a$ and $x = a$?

Edit: A condensed version

1) We know $f(x) = 0$ has a solution.

2) We can prove $f(x) = 0 \implies x = a$ using irreversible steps.

3) Now that we computed what $a$ actually is, by 1) we know that $x = a \implies f(x) = 0$ is true. Does it follow now that there exists another method to solve the equation using purely reversible steps?

Answer 1

This is impossible to answer in any rigorous way without having a precise definition of the sort of "steps" you want to consider. But basically, the answer is no.

It happens all the time in math that we can separately prove $P\Rightarrow Q$ and $Q\Rightarrow P$, but there is no direct way to prove $P\Leftrightarrow Q$ in both directions at once. So there is no reason to expect that just because you can prove both $f(x)=0\Rightarrow x=a$ and $x=a\Rightarrow f(x)=0$, you can prove $f(x)=0\Leftrightarrow x=a$ by a series of reversible steps. After all, any mathematical statement at all can be encoded in equations like this by defining your function $f$ appropriately.

Answer 2

$f(x) = 0 \Longrightarrow x = a$ tells you that if $x$ happens to be a root of $f$ then $x$ must be equal to $a$. In other words, the only possibility for a root of $f$ is $a$. However, this implication does not guarantee that $a$ is in fact a root. That's why you need to plug $a$ into $f$ and see whether or not you get zero.

Answer 3

You have derived a necessary condition for $f(x)=0$, namely that $x=a$. It is, however, still possible that $a$ is not actually a root of $f$, i.e. it need not be the case that $x=a \Rightarrow f(x)=0$ .

For example, $a$ might be outside of the original domain of $f$ and $f(a)$ might not even be properly defined. Imagine $f(x)=x$ where $f: [1,2] \rightarrow \mathbb{R}$ and look for $f(x)=0$.

Answer 4

[ EDIT ] Original answers remains in the edit history, but after clarifications given in the comments, the question appears to be (paraphrased):

Assuming that $f(x)=0$ has a unique solution, and provided we prove that $f(x)=0 \implies x=a\,$, is that sufficient to prove that $a$ is the unique solution, or is it still necessary to verify that $a$ does in fact satisfy $f(a)=0\,$?

The answer to this question is "the former" i.e. $f(x)=0 \implies x=a\,$ is sufficient to prove that $a$ is the unique root.

That's because any series of irreversible (but otherwise correct) operations can introduce additional roots, but can never discard a legitimate root. Therefore if $f(x)=0 \implies x \in A=\{a,b,c,\cdots\}$ then those $a,b,c,\cdots$ values are not all guaranteed to be roots of the original equation, but all the actual roots are guaranteed to be in $A$.

In the given case, however, the potential solution set $A=\{a\}$ contains a single value and, since it's been assumed that the equation has in fact one unique root, then that root must be $a$.